Showing $\mathbb{E}[\exp(\lambda X) | G] \leq \exp\left(\frac{\lambda^{2}}{2}\right)$ for a random variable $X$ and sub-sigma-algebra $\mathcal G$

Question

Can anyone help me finish off a proof that, given $\mathcal G$ a sub-$\sigma$-algebra of $\mathcal F$, $\lambda > 0$, and $X$ a random variable with $\mathbb{E}[X | \mathcal G] = 0, \mathbb{P}[|X| \leq 1] = 1$,

$$\mathbb{E}\left[\exp(\lambda X) | \mathcal G\right] \leq \exp\left(\frac{1}{2}\lambda^{2}\right)?$$

I've tried to use the fact that

$$\exp(\lambda x) \leq \frac{1-x}{2} \exp(-\lambda) + \frac{1+x}{2} \exp(\lambda)$$

for $x \in X$, since the exponential is convex, but I can't figure how to reduce it from here. It also is a bit of a particular challenge since I don't think I can necessarily assume the RV is normal given the wording of the question.

Answer 1

Since the formula $$ \exp(\lambda x) \leqslant \frac{1-x}{2} \exp(-\lambda) + \frac{1+x}{2} \exp(\lambda) $$ is valid for any $x\in [-1,1]$, we get that $$ \exp(\lambda X) \leqslant \frac{1-X}{2} \exp(-\lambda) + \frac{1+X}{2} \exp(\lambda) \mbox{ a.s..} $$ Now, take on both sides the conditional expectation with respect to $\mathcal G$ and use $\cosh(\lambda)\leqslant e^{\lambda^2/2}$.